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#226 by Joerg Arndt at Sat Aug 03 01:49:12 EDT 2024
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#225 by Michel Marcus at Sat Aug 03 01:25:55 EDT 2024
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#224 by Jason Yuen at Fri Aug 02 23:42:16 EDT 2024
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#223 by Jason Yuen at Fri Aug 02 23:41:52 EDT 2024
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| FORMULA
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a(n) = Sum_{m=0..n-2} Sum_{k=0..n-2] } abs(m-k). - Nicolas Bělohoubek, Nov 06 2022
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| STATUS
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approved
editing
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#222 by Joerg Arndt at Sat Jan 07 04:03:12 EST 2023
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#221 by Michel Marcus at Sat Jan 07 03:35:35 EST 2023
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Discussion
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| Sat Jan 07
| 04:01
| Michel Marcus: rather ? (1, n) (2, n-1) (3, n-2) ... (k, n-k+1) ... (n, 1)
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#220 by Michel Marcus at Sat Jan 07 03:35:07 EST 2023
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Discussion
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| Sat Jan 07
| 03:35
| Michel Marcus: ok for me
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#219 by Bernard Schott at Sat Jan 07 03:31:39 EST 2023
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| COMMENTS
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a(n+1) = Max_{s in S_n} Sum_{k=1..n} (k - s(k))^2 where S_n is the symmetric group of permutations of [1..n]; this maximum is obtained with the permutation s = [ = (1, n,) (2, n-1,) (3, n-2,...,2,) ... (k, n-k+1] (). (see Protat reference). - Bernard Schott, Dec 26 2022
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| STATUS
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proposed
editing
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Discussion
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| Sat Jan 07
| 03:33
| Bernard Schott: I have written the right permutation with its canonic form.
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#218 by Michel Marcus at Sat Jan 07 03:04:26 EST 2023
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#217 by Michel Marcus at Sat Jan 07 03:01:47 EST 2023
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| COMMENTS
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a(n+1) = Max_{s in S_n} Sum_{k=1..n} (k - s(k))^2 where S_n is the symmetric group of permutations of [1..n]; this maximum is obtained with the permutation s = ( = [n,n-1,n-2,...,2,1) (] (see Protat reference). - Bernard Schott, Dec 26 2022
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| STATUS
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proposed
editing
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Discussion
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| Sat Jan 07
| 03:04
| Michel Marcus: like this then ?
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